Recent content by martine80

ive solved it, by using v0x= v0 *cos(60) v0y=v0 *sin(60) x= cos(15) *25 y= sin(15)*25 and used : y=v0y*t-1/2(gt^2) ;)

ive solved it, by using v0x= v0 *cos(60) v0y=v0 *sin(60) x= cos(15) *25 y= sin(15)*25 and used : y=v0y*t-1/2(gt^2)

:rolleyes: :cry:

from the equation vy= v0y-gt, I find the time it takes : 1,15 s vy= 0 when it hits the grownd

its a slope , why 25 m?

velocity ! HELP 1:you have a lope with an angle of 15 degrees from the horizontal plane. You throw an object with an angle of 60 degrees from the horizontal plane up the slope and the object hits the ground 25m from the starting point. . What is start velocity? This is what I’ve done: Y...

V initial = 0 and Vfinal is 2GM(E)/3radius of the Earth ... Thank u for yore time , Ill let you know by tomorrow if I've solved the problem!

hehe, no its not homework, I don't have to do this ... Its voluntary I come jp with this : 1/2mv(0)^2+ G M(earth) m(object) (1/r1-1/r2)=1/2mv^2+ G M(earth) m(object) (1/r1-1/r2) Im sure its wrong :(

I know, but i toght i was suppose to use potentional energi in this ? I am not sure... I am about to give up now. As far as i know I am supose to end up with an integral I have to use Maple to solve ...

:cry: I don't know, I've sat down with a teacher and he didnt know eather , just the equation i told you about

I use : 1/2 mv^2 and Gravital Potential energy, and unerversial of gravity = Gm(1)m(2)/r^2

how do i do that? so far I've come up with this: 2*R(radius of the earth)=3R-integrate from 0 to t(1/2(integrate from 3R to 2R(-G*M(Earth)/r^2)) *t^2

Hi, I am a student from norway, and I am strugeling with this problem, hope someone can help me og guide me in the direction: A object is brought out in space in a distance of 3(6,37*10^6) from the senter of the earth. not any velocity when the object is released in space. 1) What is the...